. Find the dimensions of the rectangle of perimeter 36 cm which will sweep out a volume as large as possible, when revolved about one of its sides. Also find the maximum volume.

Dear student

Let’s consider x and y to be the length and breadth of given rectangle ABCD.
According to the question, the rectangle will be resolved about side AD which making a cylinder with radius x and height y.
So, the volume of the cylinder V =π r^2 h =π x^2 y.... (1)
Now, perimeter of rectangle P = 2(x + y)

36 = 2(x + y)
18 = x + y
y = 18–x..... (ii)

Putting the value of y in the equation (i),
V =π x^2 (18–x)
=π(18x^2–x^3)

Differentiating both sides w.r.t. x
dV/dx =π(36x–3x^2)................ (iii)

For local maxima and local minima dV/dx = π(36x–3x^2) = 0

36x–3x^2=0
3x(12 - x) = 0
x≠0 and 12–x = 0⇒x = 12

From equation (ii), we have
y = 18–12 = 6
Differentiating equation (iii) w.r.t. x, we get

d^2V/dx^2= π(36–6x)
At x = 12,

d^2 V/dx^2=π(36–6 x 12) =π(36–72) = -36π< 0 maxima

Now, volume of the cylinder so formed =π x^2 y
=πx (12)^2x (6)
=π(144)^2x 6
= 864πcm^3

Therefore, the required dimension are 12 cm and 6 cm and the maximum volume is 864πcm

Thanks